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 How it looks (i.θ)ₐ sheldonison Long Time Fellow Posts: 640 Threads: 22 Joined: Oct 2008 04/19/2015, 02:40 PM (This post was last modified: 04/19/2015, 03:16 PM by sheldonison.) (04/18/2015, 11:20 PM)marraco Wrote: $\vspace{15}{e^{i.\theta}}$ is an interesting and important function, so what we get if we do the same with tetration: $\vspace{15}{^{i.\theta}a}$? Your graph for base(e), and the other bases<1 is not correct. The general form for the unique slog for bipolar tetration, which is real valued and analytic, from both fixed points, for bases $b>\exp(1/e)$ is: $\text{slog}(z) = \alpha(z) +\frac{\ln(z-L)}{\ln(L\cdot \ln(b))} + \frac{\ln(z-L^{*})}{\ln(L^{*}\cdot \ln(b))}\;\;b^L=L\;\;$ This is also the form for Jay's accelerated slog Here; $\alpha(z)$ is analytic and bounded on a sickle including the fixed points. slog(z) has a singularity at the fixed points themselves, but is bounded between the fixed points, and is analytic, and has an inverse, sexp(z), on that sickle. Here, at L, $\alpha(L)$ is a constant. off topic: actually, $\alpha(z)$ has a complex singularity at z=L, but it is continuous. In the upper half of the complex plane, as $\Im(z)$ gets arbitrarily large, the sexp(z) approaches the Koenig's solution; where the exponential term below goes to zero, and the sexp(z) approaches L. In the lower half of the complex plane, it approaches the conjugate. This is the basis for Kouznetsov's solution. $\text{sexp}(z) \approx L + (L\cdot\ln(b))^{(z+k)}$ So then here is a graph of sexp(z) base e, from -5i to +5i, showing the sexp(z) function going from L* to L. You can generate sexp(z) for any complex(z) for many real bases $b>\eta$ using my implementation of Kneser's solution     - Sheldon « Next Oldest | Next Newest »

 Messages In This Thread How it looks (i.θ)ₐ - by marraco - 04/18/2015, 11:20 PM RE: How it looks (i.θ)ₐ - by sheldonison - 04/19/2015, 02:40 PM RE: How it looks (i.θ)ₐ - by marraco - 04/19/2015, 08:40 PM RE: How it looks (i.θ)ₐ - by sheldonison - 04/19/2015, 11:19 PM RE: How it looks (i.θ)ₐ - by marraco - 04/20/2015, 02:35 AM RE: How it looks (i.θ)ₐ - by Gottfried - 04/20/2015, 07:53 AM RE: How it looks (i.θ)ₐ - by marraco - 04/21/2015, 02:34 AM RE: How it looks (i.θ)ₐ - by marraco - 04/21/2015, 03:20 AM RE: How it looks (i.θ)ₐ - by sheldonison - 04/21/2015, 08:23 AM RE: How it looks (i.θ)ₐ - by marraco - 04/23/2015, 04:52 PM RE: How it looks (i.θ)ₐ - by JmsNxn - 04/23/2015, 11:15 PM RE: How it looks (i.θ)ₐ - by sheldonison - 04/23/2015, 11:20 PM RE: How it looks (i.θ)ₐ - by marraco - 04/26/2015, 12:50 AM RE: How it looks (i.θ)ₐ - by sheldonison - 04/26/2015, 05:08 AM RE: How it looks (i.θ)ₐ - by marraco - 04/20/2015, 03:46 AM RE: How it looks (i.θ)ₐ - by marraco - 04/23/2015, 02:21 AM RE: How it looks (i.θ)ₐ - by marraco - 04/25/2015, 07:52 PM

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